please respond to these two comments on my post. how would your response to what they said on my post.

For Scenario 1,

I will choose to compute a Z or a T value.

Z value because population std deviation information is given, and the underlying population is also normally distributed.

1. Compute the Z or T value for each one of your samples. Is either of these samples concerning to us? If so, what reasons may attribute to something like this?

Sample 1: Z = (31.8-29.6)/(7.5/sqrt (30)) = 1.607

Sample 2: Z = (31.6-29.6)/(7.5/sqrt (50)) = 1.885 ;

None of these is concerning because the z value is not too extreme. (z<2 for both cases)

For Scenario 2,

I Choose the T value because the population std deviation is not known, and the population is known to be normally distributed; hence cannot use the Z value.

Compute the Z or T value for each one of your samples. Is either of these samples concerning to us? If so, what reasons may attribute to something like this?

Sample 1: T value = (27.6-29.6)/ (6.8/sqrt (30)) = -1.6109

Sample 2: T value = (28-29.6)/ (7.7/sqrt (50)) = -1.4693,

None of these is concerning because the t value is less than 2 in magnitude in both cases hence values are not too extreme.

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please reply on response 1 and response 2 individually please please thank you

response 1.

hello, I got the same answers for my discussion. I liked the way that you laid out these equations to show your work in a way while laying out your answers. Though these scores are all in all non-concerning, sample 2 in scenario 1 is the closest to being concerning because it is so much closer to that 2 value of deviation. In sample 1 it shows that these samples did fall above average, however in sample 2 these samples proved to be below average due to the negative value of the distribution scale.

response 2

Hello ,

I agree that the first scenario is a Z value because it gives us the standard deviation. On the second scenario, the standard deviation was unknown which is why I also went with the T value. I also think that since both of the measurements fall within the 2 standard deviation away from the mean, the data point would not be considered extreme and would not need to be investigated.